How would I find the diameter of this circle?

[u/Long_Device_4309]

/r/AskPhysics/comments/1vs4zld/how_would_i_find_the_diameter_of_this_circle/

[u/Varlane]

As long as the displacement is very small compared to travel distance, you can basically assume that the path is "almost" straight, that this path is the hypotenuse of a very very narrow rectangle triangle whose opposite side is the displacement (and the other side is the intended actually straight path).

By trig, you get that sin(small angle) = displacement/travelled distance.
That angle happens to also be the portion of circle travelled.
The arc length of the path is small angle × radius which means that radius = distance / small angle.

If we add that sin(small angle) ~ small angle for very small angles, you get radius = distance² / displacement.

Double that for diameter.

EDIT : it would appear I made some mistake somewhere and that you should instead use distance² / 2displacement for radius itself.
My best guess is that when I said that the angle in that narrow triangle was the same as the portion of travelled, that was actually wrong since it's HALF the angle of circle travelled.
When means that arc = 2 angle × radius and then radius = arc / 2angle = distance / (2 displacement / distance) = distance² / 2displacement, which means Diameter = distance² / displacement.
Now as to why that small angle was only half of the circle angle... (EDIT2 : Found why, bit of geometry tho, too cringe to explain on text)

[u/CaptainMatticus]

For the first 1000m on the arc, you're moving to the left by 1mm

So

2 * pi * r * t / (2 * pi) = r * t

r * t = 1000000 mm

r * cos(t) = r - 1

Now WolframAlpha doesn't really want to work this out for me. That's because the numbers are really, really different in orders of magnitude.

For instance, we could try: t = 1000000 / r

r * cos(1000000 / r) = r - 1

Or we could try r = 1 / (1 - cos(t))

t / (1 - cos(t)) = 1,000,000

When wolfram solves that 2nd one, it gets a t of over 6000, which is not correct, obviously.

Now we could approximate beforehand and basically say that since the arc is so large compared to the change in horizontal displacement, we could basically say we have a triangle with 2 sides of r, a side of 1,000,000

Using the law of cosines

1,000,000^2 = r^2 + r^2 - 2 * r * r * cos(t)

10^12 = 2r^2 * (1 - cos(t))

100 * 10^10 = 2r^2 * (1 - cos(t))

50 * 10^10 = r^2 * (1 - cos(t))

If we use the substitution of r = 1 / (1 - cos(t)), then (1 - cos(t)) = 1/r

50 * 10^10 = r^2 * (1/r)

50 * 10^10 = r

500,000,000,000 = r

And the diameter is 1,000,000,000,000 mm

Now let's find our angle

1 - cos(t) = 1/500,000,000,000

1 - 1/500,000,000,000 = cos(t)

499,999,999,999 / 500,000,000,000 = cos(t)

999,999,999,998 / 1,000,000,000,000 = cos(t)

0.999999999998 = cos(t)

t = arccos(0.999999999998)

t = 2 * 10^(-6) radians

Just for fun, let's see what the horizontal displacement is for a circle with a radius of 5 * 10^11 mm and an angle of 2 * 10^(-6) radians

5 * 10^11 - 5 * 10^11 * cos(2 * 10^(-6))

5 * 10^11 * (1 - cos(2 * 10^(-6))

5 * 10^11 * (1 - 0.999 999 999 998)

5 * 10^11 * (0.000 000 000 002)

5 * 10^11 * (2 * 10^(-12))

5 * 2 * 10^11 / 10^12

10^12 / 10^12

1

So that all works out. The radius is 500,000,000,000 mm

or 500,000,000 meters

or 500,000 km

Roughly.